Question

A 32.44 mL sample of 0.202M acetic acid is titrated with 0.185 M sodium hydroxide. Calculate the pH of the solution 1. before any NaOH is added 2. after 24.00 mL of NaOH is added 3. at the equivalence point

Answer 1

CH₃COOH (aq.) <--------> CH₃COO^- (aq.) + H⁺ (aq.)

(1) Iitial mol of acetic acid = 0.202 * 32.44 / 1000 = 0.00655 mol

[H+] = sqrt.(Ka*C) = sqrt.(1.75*10-5*0.00655) = 0.000338 M

pH = - Log[H+] = - Log (0.000338) = 3.47

(2) moles of NaOH added = 0.185 * 24.00 / 1000 = 0.00444 mol

Final moles of acid = 0.00655 - 0.00444 = 0.00211 mol

moles of salt formed = 0.00444 mol

Therefore,

pH = pKa + Log[salt]/[acid]

pH = 4.75 + log(0.00444/0.00211)

pH = 5.07

(3) at equivalence point,

moles of NaOH added = moles of salt formed = 0.00655 mol

CH₃COO^- (aq.) <-----------> CH₃COOH (aq.) + OH^- (aq.)

[OH-] = sqrt.(Kb*C) = sqrt.(5.56*10-10*0.00655/0.06786) = 7.33 * 10^-6 M

pH = 14 + log[7.33*10^-6)

pH = 8.86